A manufacturer produces two Models of bikes-Model X and Model Y.Model X takes a 6 man-hours to make per unit,while Model Y takes 10 man-hours per unit.There is a total of 450 man-hour available per week.Handling and marketing costs are Rs 2000 and Rs 1000 per unit for models X and Y respectively.The total funds available for these purposes are Rs 80,000 per week.Profits per unit for models X and Y are Rs 1000 and Rs 500,respectively.How many bikes of each model should the manufacturer produce so as to yield a maximum profit.Find the maximum profit.

1 Answer

Toolbox:

Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.

If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R

Step 1:

Model of bikes$\quad$X$\;\;\;\;\quad$Y

Hours/unit$\;\;\;\quad\;\;\;$6$\;\;\;\;\quad$10

Cost/unit$\quad\;\;$Rs2000$\quad$Rs1000

Profit/unit$\quad\;\;$Rs1000$\quad$Rs500

It is given model $x$ takes 6-man-hours to make per unit and model $Y$ takes 10 man hours per unit.

No of hours available pr week is 450

$6x+10y\leq 450$

Handling and marketing costs are RS.2000 for model $X$ and Rs 1000 for model $Y$

The total funds available is Rs80,000 per week.

Therefore $2000x+1000y\leq 80,000\Rightarrow 2x+y\leq 80$

Profits per unit of models $X$ and $Y$ are Rs 1000 and Rs.500.

Hence the objective function to be maximized is $Z=1000x+500y$

Step 2:

Now let us draw the graph for the line AB :$6x+10y=450$ and CD :$2x+y=80$